Simplify the following expression: $y = \dfrac{-4x^2- 1x+5}{x - 1}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-4)}{(5)} &=& -20 \\ {a} + {b} &=& &=& {-1} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-20$ and add them together. Remember, since $-20$ is negative, one of the factors must be negative. The factors that add up to ${-1}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-5}$ and ${b}$ is ${4}$ $ \begin{eqnarray} {ab} &=& ({-5})({4}) &=& -20 \\ {a} + {b} &=& {-5} + {4} &=& -1 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-4}x^2 {-5}x) + ({4}x +{5}) $ Factor out the common factors: $ x(-4x - 5) - 1(-4x - 5)$ Now factor out $(-4x - 5)$ $ (-4x - 5)(x - 1)$ The original expression can therefore be written: $ \dfrac{(-4x - 5)(x - 1)}{x - 1}$ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ This leaves us with $-4x - 5; x \neq 1$.